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The domain of the function \(f\) defined by \(f(x)=\frac{1}{\sqrt{x-|x|}}\) is
The correct answer is (d). Given that \(f(x)=\frac{1}{\sqrt{x-|x|}}\)
where
\(
x-|x|=\left\{\begin{array}{lll}
x-x=0 & \text { if } & x \geq 0 \\
2 x & \text { if } & x<0
\end{array}\right.
\)
Thus \(\frac{1}{\sqrt{x-|x|}}\) is not defined for any \(x \in \mathbf{R}\).
Hence \(f\) is not defined for any \(x \in \mathbf{R}\), i.e. Domain of \(f\) is none of the given options.
If \(f(x)=x^{3}-\frac{1}{x^{3}}\), then \(f(x)+f\left(\frac{1}{x}\right)\) is equal to
The correct choice is \(\mathrm{c}\).
Since
\(
\begin{aligned}
f(x) &=x^{3}-\frac{1}{x^{3}} \\
f\left(\frac{1}{x}\right) &=\frac{1}{x^{3}}-\frac{1}{\frac{1}{x^{3}}}=\frac{1}{x^{3}}-x^{3}
\end{aligned}
\)
Hence,
\(
f(x)+f\left(\frac{1}{x}\right)=x^{3}-\frac{1}{x^{3}}+\frac{1}{x^{3}}-x^{3}=0
\)
Let \(\mathrm{A}\) and \(\mathrm{B}\) be any two sets such that \(n(\mathrm{~B})=p, n(\mathrm{~A})=q\) then the total number of functions \(f: \mathrm{A} \rightarrow \mathrm{B}\) is equal to
(a) Any element of set A, say \(x_{i}\) can be connected with the element of set B in \(p\) ways. Hence, there are exactly \(p^{q}\) functions.
Let \(f\) and \(g\) be two functions given by
\(
f=\{(2,4),(5,6),(8,-1),(10,-3)\}
\)
\(g=\{(2,5),(7,1),(8,4),(10,13),(11,-5)\}\) then. Domain of \(f+g\) is
(c) Since Domain of \(f=\mathrm{D}_{f}=\{2,5,8,10\}\) and Domain of \(g=\mathrm{D}_{g}=\{2,7,8,10,11\}\), therefore the domain of \(f+g=\left\{x \mid x \in \mathrm{D}_{f} \cap \mathrm{D}_{g}\right\}=\{2,8,10\}\)
Let \(n(\mathrm{~A})=m\), and \(n(\mathrm{~B})=n\). Then the total number of non-empty relations that can be defined from \(A\) to \(B\) is
The correct option is \(\mathrm{d}, 2^{\mathrm{mn}}-1\)
Let \(n(A)=m\) and \(n(B)=n\).
We have \(A \times B=\{(a, b): a \in A, b \in B\}\)
\(
\Rightarrow \mathrm{n}(A \times B)=\mathrm{n}(A) \times n(B)=m n
\)
A relation from \(A\) to \(B\) is a subset of \(A \times B\).
Since \(A \times B\) has \(m n\) elements, it has \(2^{\mathrm{mn}}\) subsets.
Thus, there can be \(2^{\mathrm{mn}}\) relations that can be defined from \(A\) to \(B\).
\(\Rightarrow\) The total number of non-empty relations (excluding the subset \(\phi\) of \(A \times B\) ) that can be defined from \(A\) to \(B\) is \(2^{m n}-1\).
If \([x]^{2}-5[x]+6=0\), where \([.]\) denote the greatest integer function, then
(d) we have
\(
[x]^{2}-5[x]+6=0
\)
We will split the middle term, we get
\(
\begin{aligned}
&\Rightarrow[\mathrm{x}]^{2}-3[\mathrm{x}]-2[\mathrm{x}]+6=0 \\
&\Rightarrow[\mathrm{x}]([\mathrm{x}]-3)-2([\mathrm{x}]-3)=0 \\
&\Rightarrow([\mathrm{x}]-3)([\mathrm{x}]-2)=0 \\
&\Rightarrow[\mathrm{x}]-3=0 \text { or }[\mathrm{x}]-2=0 \\
&\Rightarrow[\mathrm{x}]=3 \text { ог }[\mathrm{x}]=2 \\
&\Rightarrow[\mathrm{x}]=2,3
\end{aligned}
\)
For \([\mathrm{x}]=2, \mathrm{x} \in[2,3)\)
For \([\mathrm{x}]=3, \mathrm{x} \in[3,4)\)
\([x] \in[2,3) \cup[3,4)\)
So, \(x \in[2,4]\)
Range of \(f(x)=\frac{1}{1-2 \cos x}\) is
(b) We know that, \(-1 \leq-\cos x \leq 1 \Rightarrow-2 \leq-2 \cos x \leq 2\)
\(
\begin{aligned}
&\Rightarrow 1-2 \leq 1-2 \cos x \leq 1+2 \\
&\Rightarrow-1 \leq 1-2 \cos x \leq 3 \\
&\Rightarrow-1 \leq \frac{1}{1-2 \operatorname{cos} x} \leq \frac{1}{3} \\
&\Rightarrow-1 \leq f(x) \leq \frac{1}{3} \therefore \text { Range of } f=\left[-1, \frac{1}{3}\right]
\end{aligned}
\)
Let \(f(x)=\sqrt{1+x^{2}}\), then
\(\left[\right.\) Hint: find \(\left.f(x y)=\sqrt{1+x^{2} y^{2}}, f(x).f(y)=\sqrt{1+x^{2} y^{2}+x^{2}+y^{2}}\right]\)
First we will find \(f(x y)\), for this we will replace \(x\) with \(x y\) in the given equation, we get
\(
\mathrm{f}(\mathrm{xy})=\sqrt{1+(\mathrm{xy})^{2}}
\)
\(
\mathrm{f}(\mathrm{xy})=\sqrt{1+\mathrm{x}^{2} \mathrm{y}^{2}} \ldots \ldots \text { (i) }
\)
Now we will find \(f(y)\), for this we will replace \(x\) with \(y\) in the given equation, we get
\(
\mathrm{f}(\mathrm{y})=\sqrt{1+\mathrm{y}^{2}}
\)
Using this we will find the value for \(f(x) . f(y)\), we get
\(
f(x) \cdot f(y)=\sqrt{1+x^{2}} \cdot \sqrt{1+y^{2}}
\)
\(
=\sqrt{\left(1+\mathrm{x}^{2}\right)\left(1+\mathrm{y}^{2}\right)}
\)
\(
=\sqrt{1+\mathrm{x}^{2}+\mathrm{y}^{2}+\mathrm{x}^{2} \mathrm{y}^{2}} \ldots \ldots \text { (ii) }
\)
Comparing equations (i) and (ii), we get
\(
\begin{aligned}
&\sqrt{1+x^{2} y^{2}} \leq \sqrt{1+x^{2}+y^{2}+x^{2} y^{2}} \\
&\Rightarrow f(x y) \leq f(x) f(y)
\end{aligned}
\)
So, the correct answer is option (c)
Domain of \(\sqrt{a^{2}-x^{2}}(a>0)\) is
So the domain of a function consists of all the first elements of all the ordered pairs, i.e., \(x\), so we have to find the values of \(x\) to get the required domain
let,
\(
f(x)=\sqrt{a^{2}-x^{2}}(a>0)
\)
Now for real value
\(
\begin{aligned}
&a^{2}-x^{2} \geq 0 \\
&\Rightarrow x^{2} \geq a^{2} \\
&\Rightarrow x \geq \pm a
\end{aligned}
\)
Or \(-a \leq x \leq a\)
Hence the domain of given function is \(=[-a, a]\)
So, the correct answer Is option (b)
If \(f(x)=a x+b\), where \(a\) and \(b\) are integers, \(f(-1)=-5\) and \(f(3)=3\), then \(a\) and \(b\) are equal to
(b) We have, \(f(x)=a x+b\)
\(
\begin{array}{ll}
\therefore & f(-1)=a(-1)+b \\
\Rightarrow & -5=-a+b \\
\text { Also, } & f(3)=a(3)+b \\
\Rightarrow & 3=3 a+b
\end{array}
\)
On solving Eqs. (i) and (ii), we get \(a=2\) and \(b=-3\)
The domain of the function \(f\) defined by \(f(x)=\sqrt{4-x}+\frac{1}{\sqrt{x^{2}-1}}\) is equal to
So the domain of a function consists of all the first elements of all the ordered pairs, i.e., \(x\), so we have to find the values of \(x\) to get the required domain
We have
\(
f(x)=\sqrt{4-x}+\frac{1}{\sqrt{\left(x^{2}-1\right)}}
\)
Now for real value
\(4-x \geq 0\) and \(x^{2}-1>0\)
\(\Rightarrow 4 \geq x\) and \(x^{2}>1\)
\(\Rightarrow x \leq 4\) and \(-1>x>1\)
\(\Rightarrow x \leq 4\) and \(x>1\) and \(x<-1\)
\(\Rightarrow \mathrm{x} \in(-\infty,-1) \cup(1,4]\)
Hence the domain of given function is \((-\infty,-1) \cup(1,4]\)
So, the correct answer is option (a)
The domain and range of the real function \(f\) defined by \(f(x)=\frac{4-x}{x-4}\) is given by
(c) iven that: \(\mathrm{f}(\mathrm{x})=\frac{4-x}{x-4}\)
We know that \(f(x)\) is defined if \(x-4 \neq 0\) \(\Rightarrow \mathrm{x} \neq 4\)
So, the domain of \(f(x)\) is \(=R-\{4\}\)
Let \(f(\mathrm{x})=\mathrm{y}=\frac{4-x}{x-4}\)
\(
\begin{aligned}
&\Rightarrow \mathrm{yx}-4 \mathrm{y}=4-\mathrm{x} \\
&\Rightarrow \mathrm{yx}+\mathrm{x}=4 \mathrm{y}+4 \\
&\Rightarrow \mathrm{x}(\mathrm{y}+1)=4 \mathrm{y}+4 \\
&\Rightarrow \mathrm{x}=\frac{4(1+y)}{1+y}
\end{aligned}
\)
If \(x\) is real number, then \(1+y \neq 0\) \(\Rightarrow y \neq-1\)
\(\therefore\) Range of \(f(x)=R-\{-1\}\)
The domain and range of real function \(f\) defined by \(f(x)=\sqrt{x-1}\) is given by
(c) Given, \(f(x)=\sqrt{x-1}\)
\(f(x)\) is defined when \(x-1 \geq 0\)
\(\Rightarrow x \geq 1\)
\(\therefore D_{f}=[1, \infty)\)
For \(x \geq 1\), we have \(x-1 \geq 0\)
\(\Rightarrow \sqrt{x-1} \geq 0\)
\(\Rightarrow f(x) \geq 0\)
\(\therefore f(x)\) can take all real values greater than or equal to 0 .
\(\therefore R_{f}=[0, \infty)\)
The domain of the function \(f\) given by \(f(x)=\frac{x^{2}+2 x+1}{x^{2}-x-6}\)
(a) Given that: \(f(\mathrm{x})=\frac{x^{2}+2 x+1}{x^{2}-x-6}\)
\(f(x)\) is defined if \(x^{2}-x-6 \neq 0\)
\(
\begin{aligned}
&\Rightarrow x^{2}-3 x+2 x-6 \neq 0 \\
&\Rightarrow(x-3)(x+2) \neq 0 \\
&\Rightarrow x \neq-2, x \neq 3
\end{aligned}
\)
So, the domain of \(f(x)=R-\{-2,3\}\)
The domain and range of the function \(f\) given by \(f(x)=2-|x-5|\) is
(b) Given \(f(x)=2-|x-5|\)
Domain of \(f(x)\) is defined for all real values of \(x\).
Since, \(|x-5| \geq 0\)
\(
\begin{aligned}
&\Rightarrow-|x-5| \leq 0 \\
&\Rightarrow 2-|x-5| \leq 2 \\
&\Rightarrow f(x) \leq 2
\end{aligned}
\)
Hence, range of \(f(x)\) is \((-\infty, 2]\).
The domain for which the functions defined by \(f(x)=3 x^{2}-1\) and \(g(x)=3+x\) are equal is
(a) Given that: \(f(x)=3 x^{2}-1\) and \(g(x)=3+x\)
\(
\begin{aligned}
&f(x)=g(x) \\
&\Rightarrow 3 x^{2}-1=3+x \\
&\Rightarrow 3 x^{2}-x-4=0 \\
&\Rightarrow 3 x^{2}-4 x+3 x-4=0 \\
&\Rightarrow x(3 x-4)+1(3 x-4)=0 \\
&\Rightarrow(x+1)(3 x-4)=0 \\
&\Rightarrow x+1=0 \text { or } 3 x-4=0 \\
&\Rightarrow x=-1 \text { or } x=\frac{4}{3} \\
&\therefore \text { Domain }=\left\{-1, \frac{4}{3}\right\}
\end{aligned}
\)
Let \(f\) and \(g\) be two real functions given by
\(f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\}\)
\(g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}\)
then the domain of \(f . g\) is given by ____
(a) It is given that \(f\) and \(g\) are two real functions such that
\(
f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\}
\)
and \(g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}\)
Now,
Domain of \(f=D_{f}=\{0,2,3,4,5\}\)
Domain of \(g=D_{g}=\{1,2,3,4,5\}\)
\(\therefore\) Domain of \(f g=D_{f} \cap D_{g}=\{2,3,4,5\}\)
Let \(f=\{(2,4),(5,6),(8,-1),(10,-3)\}\)
\(g=\{(2,5),(7,1),(8,4),(10,13),(11,5)\}\)
be two real functions. Then Match the following:
\(
\begin{array}{|l|l|}
\hline \text { List 1 } & \text { List 2 } \\
\hline f-g & \left\{\left(2, \frac{4}{5}\right),\left(8, \frac{1}{4}\right),\left(10, \frac{-3}{13}\right)\right\} \\
\hline f+g & \{(2,20),(8,-4),(10,-39)\} \\
\hline f . g & \{(2,-1),(8,-5),(10,-16)\} \\
\hline \frac{f}{g} & \{(2,9),(8,3),(10,10)\} \\
\hline
\end{array}
\)
\(
\text { (a) } \leftrightarrow \text { (iii) } \text { (b) } \leftrightarrow \text { (iv) } \text { (c) } \leftrightarrow \text { (ii) } \text { (d) } \leftrightarrow \text { (i) }
\)
State True or False for the following statements:
\(\text { The ordered pair }(5,2) \text { belongs to the relation } \mathrm{R}=\{(x, y): y=x-5, x, y \in \mathbf{Z}\}\)
The given statement Is false.
Explanation:
given \(R=\{(x, y): y=x-5, x, y \in \mathbf{Z}\}\)
This means set \(\mathrm{R}\) contains numbers such that \(y=x-5\), so
So when \(x=5, y\) becomes
\(
y=x-5=5-5=0
\)
So corresponding y will be 0 ,
So \((5,2)\) does not belong to \(R\).
State True or False for the following statements:
\(\text { If } P=\{1,2\} \text {, then } P \times P \times P=\{(1,1,1),(2,2,2),(1,2,2),(2,1,1)\}\)
False
We have, \(P=\{1,2\}\) and \(n(P)=2\)
\(
n(P \times P \times P)=n(P) \times n(P) \times n(P)=2 \times 2 \times 2
\)
\(=8, \mathrm{But}\) given \(\mathrm{P} \times \mathrm{P} \times \mathrm{P}\) has 4 elements.
State True or False for the following statements:
If \(\mathrm{A}=\{1,2,3\}, \mathrm{B}=\{3,4\}\) and \(\mathrm{C}=\{4,5,6\}\), then \((\mathrm{A} \times \mathrm{B}) \cup(\mathrm{A} \times \mathrm{C})\)
\(=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5),(3,6)\}\)
True
We have. \(4=\{1,2,3\}, 5=\{3,4\}\) and \(C=\{4,5,6\}\)
\(A \times B=\{(1,3),(1,4),(2,3),(2,4),(3,3),(3,4)\}\)
and \(A \times C=\{(1,4),(1,5),(1,6),(2,4),(2,5),(2,6),(3,4),(3,5),(3,6)\}\)
\(
\begin{aligned}
&(A \times B) \cup(A \times C)=\{(1,3),(1,4),(1,5),(1,6),(2,3),(2,4),(2,5),(2,6),(3,3),(3,4),(3,5), \\
&(3,6)\}
\end{aligned}
\)
State True or False for the following statements:
\(
\text { If }(x-2, y+5)=\left(-2, \frac{1}{3}\right) \text { are two equal ordered pairs, then } x=4, y=\frac{-14}{3}
\)
The given statement is false.
Explanation:
Two ordered pairs are equal, if and only if the corresponding first elements are equal and the second elements are also equal,
So from given criteria
\(
(\mathrm{x}-2, \mathrm{y}+5)=\left(-2, \frac{1}{3}\right)
\)
\(\Rightarrow \mathrm{x}-2=-2\) and \(\mathrm{y}+5=\frac{1}{3}\)
\(
\Rightarrow \mathrm{x}=-2+2 \text { and } \mathrm{y}=\frac{1}{3}-5
\)
\(
\Rightarrow \mathrm{x}=0 \text { and } \mathrm{y}=\frac{1-15}{3}=-\frac{14}{3}
\)
These are the values of \(x\) and \(y\).
State True or False for the following statements:
\(
\text { If } \mathrm{A} \times \mathrm{B}=\{(a, x),(a, y),(b, x),(b, y)\}, \text { then } \mathrm{A}=\{a, b\}, \mathrm{B}=\{x, y\}
\)
It is given that \(\mathrm{A} \times \mathrm{B}=\{(a, x),(a, y),(b, x),(b, y)\}\)
We know that the Cartesian product of two non-empty sets \(P\) and \(Q\) is defined as \(P \times Q=\{(p, q)\) : \(\left.p \in \mathrm{P}, q \in \mathrm{Q}\right\}\)
\(\therefore A\) is the set of all first elements and \(B\) is the set of all second elements.
Thus, \(\mathrm{A}=\{a, b\}\) and \(\mathrm{B}=\{x, y\}\)
\(
\text { If } f(x)=y=\frac{a x-b}{c x-a} \text {, then prove that } f(y)=?\)
(a)
\(
f(x)=y=\frac{a x-b}{c x-a}
\)
Express \(x\) in terms of \(y\)
\(
y(c x-a)=a x-b
\)
\(
c x y-a y=a x-b
\)
\(
c x y-a x=a y-b
\)
\(
x(c y-a)=a y-b
\)
\(
\mathrm{x}=\frac{\mathrm{ay}-\mathrm{b}}{\mathrm{cy}-\mathrm{a}}=\mathrm{f}(\mathrm{y})
\)
\(\text { What is the domain and Range of the function } f(x)=\frac{1}{\sqrt{x-5}}?\)
(d) We have, \(f(x)=\frac{1}{\sqrt{x-5}}\)
\(\mathrm{f}(\mathrm{x})\) is defined, if \(x-5>0 \Rightarrow x>5\)
\(\therefore\) Domain of \(f=(5, \infty)\)
Let \(f(x)=y\) )
\(\therefore y=\frac{1}{\sqrt{x-5}} \Rightarrow \sqrt{x-5}=\frac{1}{y}\) \(\Rightarrow x-5=\frac{1}{y^{2}}\) \(\therefore x=\frac{1}{y^{2}}+5\) \(\therefore x \in(5, \infty) \Rightarrow y \in R^{+}\) Hence, range of \(f=R^{+}\)
If \(f(x)=\frac{x-1}{x+1}\), then what is
(i) \(f\left(\frac{1}{x}\right)\)?
(ii) \(f\left(-\frac{1}{x}\right)\)?
(a) Given that \(f(x)=(x-1) /(x+1)\)
(l) \(f(1 / x)=-f(x)\)
\(
f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{1-x}{1+x}=\frac{-(x-1)}{x+1}=-f(x)
\)
Hence \(f\left(\frac{1}{x}\right)=-f(x)\)
(II) \(f(-1 / x)=-1 / F(x)\)
\(
f\left(\frac{-1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{-\left(\frac{1}{x}+1\right)}{-\left(\frac{1}{x}-1\right)}=\frac{1+x}{1-x}
\)
\(
=\frac{1}{\frac{1-x}{1+x}}=\frac{1}{\left(\frac{x-1}{x+1}\right)}=\frac{-1}{f(x)}
\)
Hence, \(f\left(\frac{-1}{x}\right)=\frac{-1}{f(x)}\)
Let \(f(x)=\sqrt{x}\) and \(g(x)=x\) be two functions defined in the domain \(\mathrm{R}^{+} \cup\{0\}\). Find the value of the following items
(i) \((f+g)(x)\)
(ii) \((f-g)(x)\)
(iii) \((f g)(x)\)
(iv) \(\left(\frac{f}{g}\right)(x)\)
(b) We have, \(f(x)=\sqrt{x}\) and \(g(x)=x\) be two function defined in the domain \(R^{+} \cup\{0\}\)
(i) \((f+g)(x)=f(x)+g(x)=\sqrt{x}+x\)
(ii) \((f-g)(x)=f(x)-g(x)=\sqrt{x}-x\)
(iii) \((f g)(x)=f(x) \cdot g(x)=\sqrt{x} \cdot x=x^{\frac{3}{2}}\)
(iv) \(\left(\frac{f}{g}\right)(x)=\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}}\)
\(
\text { Redefine the function } f(x)=|x-2|+|2+x|,-3 \leq x \leq 3
\)
(a) \(\begin{gathered}
-2 x,-3 \leq x<-2 \\
f(x)=4\quad,-2 \leq x<2 \\
2x,2 \leq x \leq 3
\end{gathered}\)
Find the range of the following functions given by
(i) \(f(x)=\frac{3}{2-x^{2}}\)
(ii) \(f(x)=1-|x-2|\)
(iii) \(f(x)=|x-3|\)
(iv) \(f(x)=1+3 \cos 2 x\)
We have, \(f(x)=\frac{3}{2-x^{2}}\)
Let \(y=f(x)\)
Then, \(y=\frac{3}{2-x^{2}} \Rightarrow 2-x^{2}=\frac{3}{4}\)
\(\Rightarrow x^{2}=2-\frac{3}{y} \Rightarrow x=\sqrt{\frac{2 y-3}{y}}\)
x assumes real values, if \(2 y-3 \geq 0\) and \(y>0 \Rightarrow y \geq \frac{3}{2}\)
\(\therefore\) Range of \(f=\left[\frac{3}{2}, \infty\right)\)
(ii) We known that, \(|x-2| \geq\) or \(\Rightarrow-|x-2| \leq 0\)
\(\Rightarrow 1-|x-2| \leq 1 \Rightarrow f(x) \leq 1\)
\(\therefore\) Range of \(f=(-\infty, 1]\)
(iii) We know, that, \(|x-3| \leq\) or \(\Rightarrow f(x) \geq 0\)
\(\therefore\) Rang of \(f=0, \infty]\)
(iv) We hnow that, \(-1 \leq \cos 2 x \leq 1 \Rightarrow-3 \leq 3 \cos 2 x \leq 3\)
\(\Rightarrow 1-3 \leq 1+3 \cos 2 x \leq 1+3 \Rightarrow-2 \leq 1+3 \cos 2 x \leq 1+3\)
\(\Rightarrow-2 \leq f(x) \leq 4\)
\(\therefore\) Range of \(f=[-2,4]\)
Find the domain of each of the following functions given by
(i) \(f(x)=\frac{1}{\sqrt{1-\cos x}}\)
(ii) \(f(x)=\frac{1}{\sqrt{x+|x|}}\)
(iii) \(f(x)=x|x|\)
(iv) \(f(x)=\frac{x^{3}-x+3}{x^{2}-1}\)
(v) \(f(x)=\frac{3 x}{2 x-8}\)
(i) We have \(f(x)=\frac{1}{\sqrt{1-\cos x}}\)
\(
\begin{aligned}
&\therefore -1\leq \cos x \leq 1 \\
&\Rightarrow-1 \leq-\cos x \leq 1 \\
&\Rightarrow 0 \leq 1-\cos x \leq 2
\end{aligned}
\)
So, \(\mathrm{f}(\mathrm{x})\) is defined, if \(1-\cos x \neq 0\)
\(\cos x \neq 1\)
\(x \neq 2 n \pi \forall n \in Z\)
\(\therefore\) Domain of \(f=R-\{2 n \pi: n \in Z\}\)
(ii) We have, \(f(x)=\frac{1}{\sqrt{x+|x|}}\)
\(\therefore x+|x|=x-x=0, x<0\)
\(=x+x=2 x, x \geq 0\) Hence, \(\mathrm{f}(\mathrm{x})\) is defined, if \(x>0\)
\(\therefore\) Domain of \(f=R^{+}\)
(iii) We have, \(f(x)=x (\mid x \mid)\)
clearly, \(f(x)\) is defined for any \(x \in R\).
(iv) We have, \(f(x)=\frac{x^{3}-x+3}{x^{2}-1}\)
\(\mathrm{f}(\mathrm{x})\) is not defined, If \(x^{2}-1=0\)
\(\Rightarrow(x-1)(x+1)=0\)
\(\Rightarrow x=-1,1\)
\(\therefore\) Domain of \(f=R-\{-1,1\}\)
(v) We have, \(f(x)=\frac{3 x}{2x-8}\)
Clearly, \(\mathrm{f}(\mathrm{x})\) is defined, if \(2x-8 \neq 0\)
\(\Rightarrow x \neq 4\)
\(\therefore\) Domain of \(f=R-\{4\}\)
Is \(g=\{(1,1),(2,3),(3,5),(4,7)\}\) a function? Justify. If this is described by the relation, \(g(x)=\alpha x+\beta\), then what values should be assigned to \(\alpha\) and \(\beta\)?
(c)
\(
g=\{(1,1),(2,3),(3,5),(4,7)\}
\)
Here, each element of domain has unique image. So, \(g\) is a function.
Now, given that, \(g(x)=\alpha x+\beta\)
\(
\begin{aligned}
&g(1)=1 \\
&\Rightarrow \alpha+\beta=1 \\
&g(2)=3 \\
&\Rightarrow 2 \alpha+\beta=3
\end{aligned}
\)
On solving (1) and (2), we get,
\(
\begin{aligned}
&\alpha=2, \beta=-1 \\
&\therefore g(x)=2 \mathrm{x}-1
\end{aligned}
\)
Find the values of \(x\) for which the functions \(f(x)=3 x^{2}-1\) and \(g(x)=3+x\) are equal
(d) It is given that the functions \(f(x)=3 x^{2}-1\) and \(g(x)=3+x\) are equal.
\(
\begin{aligned}
&\therefore f(x)=g(x) \\
&\Rightarrow 3 x^{2}-1=3+x \\
&\Rightarrow 3 x^{2}-x-4=0 \\
&\Rightarrow(x+1)(3 x-4)=0 \\
&\Rightarrow x+1=0 \text { or } 3 x-4=0 \\
&\Rightarrow x=-1 \text { or } x=\frac{4}{3}
\end{aligned}
\)
Hence, the set of values of \(x\) for which the given functions are equal is \(\left\{-1, \frac{4}{3}\right\}\).
\(
\text { Given } \mathrm{R}=\left\{(x, y): x, y \in \mathbf{W}, x^{2}+y^{2}=25\right\} \text {. Find the domain and Range of } \mathrm{R} \text {. }
\)
(b)
We have, \(R=\left\{(x, y): x, y \in W, x^{2}+y^{2}=25\right\}\) \(=\{(0,5),(3,4),(4,3),(5,0)\}\)
Domain of \(R=\) Set of first element of ordered pairs in \(R\) \(=\{0,3,4,5\}\)
Range of \(R=\) Set of second element of ordered pairs in \(R\) \(=\{5,4,3,0\}\)
If \(\mathrm{R}_{1}=\{(x, y) \mid y=2 x+7\), where \(x \in \mathbf{R}\) and \(-5 \leq x \leq 5\}\) is a relation. Then find the domain and Range of \(R_{1}\).
(b) We have, \(R_{1}=\{(x, y) \mid y=2 x+7\), where \(x \in R\) and \(-5 \leq x \leq 5\}\)
Domain of \(\mathrm{R}_{1}=\{-5 \leq \mathrm{x} \leq 5, \mathrm{x} \in \mathrm{R}\}=[-5,5]\)
\(
\begin{aligned}
&x \in[-5,5] \\
&\Rightarrow 2 x \in[-10,10] \\
&\Rightarrow 2 x+7 \in[-3,17]
\end{aligned}
\)
Range is \([-3,17]\)
\(\text { If } \mathrm{R}_{2}=\left\{(x, y) \mid x \text { and } y \text { are integers and } x^{2}+y^{2}=64\right\} \text { is a relation. Then find } \mathrm{R}_{2} \text {. }\)
(d) We have, \(R_{2}=\left\{(x, y) \mid x\right.\) and \(y\) are integers and \(\left.x^{2}+y^{2}=64\right\}\)
Clearly, \(x^{2}=0\) and \(y^{2}=64\) or \(x^{2}=64\) and \(y^{2}=0\)
\(x=0\) and \(y=\pm 8\)
or, \(x=\pm 8\) and \(y=0\)
\(
R=\{(0,8),(0,-8),(8,0),(-8,0)\}
\)
If \(\mathrm{R}_{3}=\{(x,|x|) \mid x\) is a real number \(\}\) is a relation. Then find domain and range of \(\mathrm{R}_{3}\).
(a)
We have, \(R_{3}=\{(x, \mid x)) \mid x\) is real number \(\}\)
Clearly, domain of \(\mathrm{R}_{3}=\mathrm{R}\)
Now, \(x \in R\) and \(|x| \geq 0\)
Range of \(R_{3}\) is \([0, \infty)\)
Let \(f\) and \(g\) be real functions defined by \(f(x)=2 x+1\) and \(g(x)=4 x-7\).
(i) For what real numbers \(x, f(x)=g(x)\)?
(ii) For what real numbers \(x, f(x)<g(x)\)?
(c) We have, \(f(x)=2 x+1\) and \(g(x)=4 x-7\)
(a) Now \(f(x)=g(x)\)
\(
\begin{aligned}
&\Rightarrow 2 x+l=4 x-7 \\
&\Rightarrow 2 x=8 \Rightarrow>x=4
\end{aligned}
\)
(b) \(f(x)<g(x)[latex]
[latex]
\begin{aligned}
&\Rightarrow 2 x+1<4 x-7 \\
&\Rightarrow 8<2 x \\
&\Rightarrow x>4
\end{aligned}
\)
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